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Expanding on Daniel's comment: suppose that $\langle x,y\rangle = \langle x,z\rangle$. Then we must have $$\langle x,y-z \rangle = \langle x,y \rangle - \langle x,z \rangle = 0$$ For all $x$. It certainly follows that, in the instance that $x = y-z$, we have $\langle x,y-z \rangle = 0$. Thus, we have $$\|y-z\| = \sqrt{\langle y-z,y-z \rangle} = 0$$ ...

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Also you can think of it this way.The only vector in the Hilbert space $\mathcal{H}$ that is perpendicular to every single vector of $\mathcal{H}$ is zero.

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Let $\delta(z) = \langle x-y,z\rangle$ and $L:\to H'$ be the Riezs bijection: $$L(a)(z) = \langle a,z\rangle$$ As $L(x-y) = \delta = 0_{H'} = L(0_H)$, $x=y$.

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In order to make the inner product well-defined, we talk about $L^2(\Omega,\mathcal F,\mu)$, where $(\Omega,\mathcal F,\mu)$ is the underlying probability space. But we then extend condition expectation to integrable random variables. We use a projection over the closed subspace $L^2(\Omega,\mathcal N,\mu)$, that is, the vector subspace which consists of ...

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This is a very interesting issue. In fact, this isomorphism is not as "natural" as one might have thought. As an exercise, one should see that, given a map of Hilbert spaces $V\to W$ it is rarely the case that the square of maps involving $W^*\to V^*$ and the "Riesz-Fisher" dualities ... commutes. This is fairly crazy, yes, given the standard curriculum. A ...

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Let $M$ be the diameter of $B$. Each of the closed sets has diameter $< M$, and is therefore contained in a closed ball that also has diameter $<M$. If the covering by arbitrary sets was finite, this would give a covering by finitely many closed balls each of whose diameters was less than $M$.

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No. Take a Hilbert space with a sequence $u_k$ such that $||u_k||=1$ while $u_k\rightharpoonup 0$. Then take $Au=u$. You have $(Au_k,u_k) = (u_k,u_k) = 1$ while $(u,u)=0$.

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Fix a basis $v_1,\ldots,v_m \in V_k$. Then the map $\sum \alpha_jv_j\to\left(\sum |\alpha_j|^2\right)^{1/2}$ defines a norm on $V_k$, and this norm is induced by the inner product $\langle \sum \alpha_jv_j,\sum \beta_jv_j\rangle = \sum \alpha_j\overline{\beta_j}$. In a finite-dimensional space all norms are equivalent, so the identity map becomes a ...

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It's easy to see that $A$ is symmetric on its dense domain, which means that $A^{\star}$ extends $A$. In order to show that $A$ is selfadjoint, it is enough to show that $(A\pm iI)$ are surjective, which is something that is trivial to demonstrate in this case. Indeed, if $f \in L^{2}$, then $(x\pm i)^{-1}f$ are in $L^{2}$ with their respective images under ...

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Since $A$ is densely defined, we can characterise $D(A^\ast)$ as \begin{align} D(A^\ast) &= \left\lbrace f \in H : g \mapsto (f, Ag) \text{ is continuous}\right\rbrace\\ &= \left\lbrace f\in H : \bigl(\exists K_f < \infty\bigr)\bigl(\forall g\in D(A)\bigr)\bigl(\lvert (f,Ag)\rvert \leqslant K_f\cdot \lVert g\rVert\bigr)\right\rbrace. ... 1 By the Lebesgue differentiation theorem,(Df)^\prime(t)=\frac{d}{dt} \int_0^t f(s)ds=f(t)$$for a.e. t\in[0,1]. Therefore,$$\langle Df, Dg\rangle_{C^\prime}=\int_0^1 (Df)^\prime(t) (Dg)^\prime(t) dt = \int_0^1 f(t) g(t) dt = \langle f,g\rangle_{L^2[0,1]}$$so D preserves the inner product. By the characterization of absolutely continuous functions by ... 2 I started writing this as comments, but then ran out of space to give a satisfactory reply. (I rant too much to be confined to 400 characters!) At any rate, the answer is: no. But some clarification is necessary. I'm old and forgetful, so I will note John Baez describes the basic algorithm to geometric quantization fairly well. Examples Worth Considering ... 3 In Banach space X a sequence \{f_n\} converges weakly to f if$$ \varphi(f_n)\to\varphi(f), $$for all \varphi\in X^*, where X^* is the dual of X. In the case of Hilbert space H, every element of the dual space is realized by an element of H (Riesz Representation Theorem). Thus f_n\to f weakly if and only if$$ \langle ...

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Let $\{f_n\}_{n=1}^\infty$ be a sequence in $H$, and let $f\in H$. Then: $f_n\to f$ in the weak topology if and only if for all $\phi\in H^*$ one has the convergence $\phi(f_n)\to \phi(f)$ as $n\to \infty$. Why is this? Let's first look at the direction ($\Rightarrow$). So we assume $f_n\to f$ in the weak topology, and want to prove $\phi(f_n)\to ... 0 If a Banach space$(X, \|\cdot\|$is a Hilbert space, then the norm satisfies the "parallelogram identity" $$\|x+y\|^2+\|x-y\|^2=2\big(\|x\|^2+\|y\|^2\big).$$ But the norm of$C[0,1]$does not satisfy such an identity: for$f=1$and$g=x$, $$\|f\|=1,\,\,\|g\|=1,\,\,\|f+g\|=2,\,\,\|f-g\|=1,$$ and hence $$... 0 Banach spaces that are not a Hilbert space are, among many others, L^{p}(Rⁿ,dⁿx) for p∈[1,∞),p≠2. Linear functionals on such spaces can be written as an integral similar to the Hilbert space inner product but in general the functional cannot be associated with an element of the space itself. But there exists the notion of a semi-scalar product which was ... 0 The space C[0,1] of continuous functions f:[0,1]\to\mathbb R with the supremum norm is an example of a Banach space which is not a Hilbert space. We need to check that the parallelogram law is not satisfied. Take f(x)=x, x\in[0,1], and g(x)=1, x\in[0,1]. Then 2(\|f\|_\infty^2+\|g\|_\infty^2)=4, but \|f+g\|_\infty^2+\|f-g\|_\infty^2=5. 0 The spectral theorem is a general argument. For a specific normal operators, there may be a concrete "diagonalizing" unitary. Take the shift S on l^2(\mathbb{Z}). In this concrete example, the diagonalizing unitary is the Fourier transform$$ \mathcal{F} : l^2(\mathbb{Z}) \rightarrow L^2(\mathbb{T}). $$The operator$$ \mathcal{F} S\mathcal{F}^{-1} ... 1 Your proof seems perfect. (Well, a small thing:$\ker(L)$is closed since$L$is continuous linear..) If$x\in K\subseteq H$, then let$y:=L(x)\,\in H'$, then by definition we have$S(y)=x$. The last part (that$S$is determined uniquely by the mentioned properties) is still to be proved. 3 Consider$H=L^2(\Omega,\mu)$where$\Omega\subseteq\mathbb C$is compact, and multiplication operator$T:H\to H,\ (Tf)(z)=zf(z).$Then$T$is bounded and normal. By the spectral theorem$T=\int_{\mathbb C}\lambda dE(\lambda).$In this case you can give an explicit formula for the spectral measure$E:$$$(E(X)f)(z)=\chi_X(z)f(z),$$ where$\chi_X$is the ... 0$T_0$is surjective cause$T$is, and injective cause you killed the kernel of T! Let$h \in H$, pick any$x \in H$such that$Tx=h$(exists by surjectivity). Write$x=x_1 + x_2 \in N \oplus N^\perp$. Then$Tx_1=0$, so$Tx_2=T_0x_2=h$, and you're done for surjectivity. Finally, if$T_0x=0$, for any$x \in N^\perp$, then$Tx=0$, so$x \in N$, and$x \in N ...

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(1): The magical ingredient that makes Hilbert spaces behave so much nicer than general Banach spaces is the parallelogram identity: Lemma: Let $H$ be a Hilbert space and $x,y\in H$. Then $$\|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2)$$ Proof: Verify using $\|x\|^2=\langle x,x\rangle$ and straightforward calculation. $\square$ Theorem: Let $H$ be a ...

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In what follows we prove that: If $\{\lambda_n\}$ are eigenvalues (corresponding to linearly independent eigenvectors) of the self-adjoint compact operator $K$, then $\lambda_n\to 0$. Suppose not, and in particular, that there exists an $\varepsilon>0$, such that $\lvert\lambda_{j_n}\rvert\ge\varepsilon$, for $\{\lambda_{j_n}\}$ a subsequence of ...

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If a product of compact sets lies within the direct sum, it will always be compact. The space in question is metrisable, so it's enough to check sequential compactness, and thanks to completeness, it's not hard to do that using the standard diagonal argument. This is not a consequence of Tychonoff's theorem, however, as Hilbert topology is much finer than ...

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Your equation is not valid, since in finite dimensions, a Hamiltonian $H$ acts on a density matrix $\rho$ via the commutator: $$[H,\rho]=H\rho-\rho H$$ Using $\rho=|V\rangle\langle V|$ and superoperator formalism i.e. $\rm{vec}(\rho)=|V\rangle ^* \otimes |V\rangle$ you could argue that: $$\hat H \rm{vec}(\rho)=(1\otimes H-H^T\otimes 1)|V\rangle ^* ... 2 f is continuous, because its bounded, so K=f^{-1}(\{0\}) is closed as preimage of a closed set under a continuous map. 1 The correct statement is that a Banach space such all of its closed subspaces are complemented is isomorphic to a Hilbert space. It was proven by Lindestrauss and Tzafriri in 1971. The proof is not very hard but not trivial either. 0 When using the Gram-Schmidt orthonormalization process, we need a linearly independent set. Is this the case here? 1 The statement in the comment is correct. The reasoning in the question is also correct on the technical level, but I can't really tell how adequate it is in the context where you'll put it. Generally, by the time people learn what a Hilbert space is, they already know how to prove that a linear subspace of \mathbb R^n is closed (because they know enough ... 3 The shortest proof I know runs as follows. If f\in A^2 and if you write f(z)=\sum_0^\infty a_n(f) z^n then, using polar coordinates and Parseval formula you find that$$\Vert f\Vert_{A^2}^2=\pi\,\sum_{n=0}^\infty \frac{\vert a_n(f)\vert^2}{n+1}\cdot$$Conversely, if (a_n)_{n\geq 0} is a sequence of complex numbers such that \sum_0^\infty ... 1 It is true if you assume that T is self-adjoint (i.e. symmetric), meaning that$$ (Tx,y)=(x,Ty), \quad \text{for all}\,\, x,y\in H, \tag{1} $$and assuming that$$ |(Tx,x)|\le \|x\|^2, \quad \text{for all}\,\, x\in H.\tag{2} $$Note that your inequality holds even for T=-2I, and thus we NEED to assume these two additional things: (1) and (2). So ... 0 The reason why a discrete topology is required for I, is to make sure that the \sigma-algebra of the Borel sets is the whole power set of I, i.e. {\mathscr P}(I). 4 Bessaga showed something stronger, but only for Hilbert spaces. Generalization to certain Banach spaces (i.e., those which are linearly injectable into some c_0(\Gamma)) was given by Dobrowolski: In particular, in 1966 C. Bessaga [1] proved that every infinite-dimensional Hilbert space H is C^\infty diffeomorphic to its unit sphere. The key to prove ... 1 Let \Phi:B(H,K)\rightarrow S(H,K) be the morphism you defined. Given \phi\in B(H,K), you can verify that \Vert\phi\Vert=\sup\left\{|\langle\phi(x),y\rangle|:x\in H, y\in K, \Vert x\Vert\leq 1,\Vert y\Vert\leq 1\right\}=\Vert\langle\phi(\cdot),\cdot\rangle\Vert=\Vert\Phi(\phi)\Vert, so \Phi is an isometry. Also, it follows (almost directly) from Riesz ... 0 When you change the norm of a normed vector space, you change the space - the second space with the sup-norm is a fundamentally different space than the space defined. The norm (and vector space) are both fundamental portions of a particular normed vector space - if you change either, you get a different space when talking about normed vector spaces (which ... 2 A L^2-Riemann Cauchy sequence of L^2-Riemann functions can have a nonintegrable limit (see the example of David Mitra). In general, The Riemann integral behaves much worse that the Lebesge integral. Example: pick a enumeration of rationals in [0,1], r_1,r_2,\cdots. The sequence$$f_n=\chi_{\{r_1,\cdots,r_n\}}$$of Riemann integrable functions has a ... 1 THIS ANSWER IS WRONG!!: Let v\in H be any non-zero vector. What can you say about the map$$\varphi\mapsto \varphi(v):B(H)\to \mathbb C?3 \forall x, y, we have following two equations: \begin{align} <T(x+iy), x+iy> = 0 \\ <T(x+y), x+y> = 0 \end{align} Since <Tx, x> = <Ty, y> = 0, these two are equivalent to \begin{align} <Ty, x> - <Tx, y> = 0 \\ <Ty, x> + <Tx, y> = 0 \end{align} Because x and y are chosen arbitrarily, we ... 2 The polynomial algebra \mathbb{C}[T,T^*] is dense in A. A polynomial p(T,T^*) is sent to the corresponding polynomial function p(z,\overline{z}) on the spectrum of T. The general description then comes from continuity. For example, when ||T|| < 1, the element \frac{1}{1-T} = \sum_{k=0}^{\infty} T^k is sent to \frac{1}{1-z} = ... 1 No matter what, the projection is still orthogonal, so it still takes the form\tilde h = P_m(h) = \sum_{j=1}^m a_jv_j$$i.e. it is in the linear span of the v_j's. If the v_j's are orthonormal, then the coefficients are "easy" to calculate:$$ a_j = \left<h,v_j\right> $$In general, the coefficients have to satisfy a system of equations. In ... 0 Note that L^p=(L^q)^*, for$$ \frac{1}{p}+\frac{1}{q}=1, $$and hence you can use the uniform boundedness principle. In particular, if \ell(f_n) is bounded for every \ell\in (L^p)^*, then you can equivalently say that f_n(\ell) is bounded for every \ell\in L^q, and consider f_n\in(L^q)^*, and hence UBP. On the other hand, it would be ... 3 This is an old joke that's done the rounds on the internet a few times. Stephan Rauh gave it a pretty good treatment on his blog last year, in which he concludes that it is indeed nonsense: Funny thing is it took me a while to figure out that the sentence really is utter nonsense. Most people immediately dismiss it as a joke – but they’ll never know ... 2 No. But it has the following universal property: H represents the contravariant functor which maps a Hilbert space K to the set of bounded families of bounded linear maps g_i : K \to H_i which are l^2-summable in the sense that \sum_{i \in I} ||g_i(x)||^2<\infty for all x \in K. Because then we can define g : K \to H by g(k)=(g_i(x)). 2 An element a of C^*-algebra is called an isometry if a^* a = 1. This coincides with the usual notion when applied to B(H). Maybe we can just generalize this terminology to the setting of dagger categories. So call a morphism f an isometry if f^{\dagger} f = 1. I don't know if this is standard. 0 The function f(x)=e^{-|x|} has a weak derivative f'(x)=-{\rm sgn\,}(x)\,e^{-|x|}, while f,f'\in L^2(\mathbb{R}). Hence f\in W^{1,2}(\mathbb{R}). But f' does not have a weak derivative, since its disribution derivative f''=e^{-|x|}+2\delta\, e^{-|x|}=e^{-|x|}+2\delta in \mathcal{D}'(\mathbb{R}). Hence f\notin ... 1 The algebra is commutative, and isomorphic to C(X) where X is the maximal ideal space. Under the Gelfand Transform, the operator itself goes to the function f(z)=z. Have a look at continuos functional calculus in your book- they should prove this statement over there. 2 The answer is yes. This is particularly easy to prove using the "multiplication operator" version of the spectral theorem, for which a good reference is Section VIII.3 of Reed and Simon. Consider the special case that \mathcal{H} = L^2(X,\mu) for some finite measure space (X,\mu), and that P is multiplication by some real-valued measurable function ... 2 Assuming the axiom of choice, yes. Let A' be the unique bounded extension. Pick a Hamel basis B for \mathcal{D} and extend it to a Hamel basis B' for \mathcal{H}. Now define \tilde{A} on B' \setminus B any way you like, as long as it's different from A'. The result will necessarily be an unbounded linear operator. 2 Note that \langle x, e_n \rangle = x_n since (e_n) forms an orthonormal basis. Similarly \langle y,e_n \rangle = y_n Then what you asked simplifies to$$ \langle x,y \rangle = \sum_{n} \langle x , e_n \rangle \langle y , e_n \rangle = \sum_{n} x_n y_n,  as you are familiar with seeing in the Euclidean case, and have likely seen as the inner product ...

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This is true even in operator space setting. See proposition 7.2 in Introduction to operator space theory. G. Pisier

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